# Synthetic Autunite reaction with HCl

The mineral autunite, $$\ce{Ca(UO2)2(PO4)2}$$, is usually present in a hydrated form (i.e., $$\ce{Ca(UO2)2(PO4)2 * {10}-{12} H2O}$$). I'm interested in understanding the reaction with hydrochloric acid. I attempted to balance the reaction equation as follows:

$$\ce{Ca(UO2)2(PO4)2 + 4 HCl -> Ca(H2PO4)2 + 2 UO2 + 2 Cl2}$$

Will this practically happen in a weak HCl solution? Would it need to be heated? Would there be any additional side-products.

• Is it pure autinite? Uranium ores typically contain only a little uranium-bearing minerals embedded in more common rock, and the surrounding rock is not in general highly soluble in HCl. Oct 19 at 10:53
• Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. Oct 19 at 12:07
• @Oscar Lanzi The autunite is synthetically produced and has high purity. Oct 19 at 14:36
• @Buttonwood Thanks for the tip. Oct 19 at 14:36

In such a reaction, the oxidation number are not modified. This dissolution is not a redox reaction. It will not produce $$\ce{Cl2}$$ and $$\ce{UO2}$$ (as stated by Young), which would imply an oxidation of Chlorine by Uranium. To find the equation, consider that autunite is made of the following ions : $$\ce{Ca(UO2)2(PO4)2 -> Ca^{2+} + 2 UO2^{2+} + 2 PO4^{3-}}$$ When adding $$\ce{HCl}$$ solution, the $$\ce{H+}$$ ions will react with the phosphate ion and it is the main change. If a very small amount of $$\ce{HCl}$$ is added, this addition will produce a precipitate of $$\ce{UO2HPO4}$$, and not $$\ce{CaHPO4}$$ as proposed as proposed by Young (there is too many phosphate ions in autunite - the only calcium phosphate which can be formed here is $$\ce{Ca(H2PO4)2}$$ - see later on). The reaction is : $$\ce{Ca(UO2)2(PO4)2 + 2 HCl -> CaCl2(aq) + 2 UO2HPO4(s)}$$ If the solution is concentrated enough, and with adding more $$\ce{HCl}$$, the following reaction may happen, remembering that $$\ce{Ca(H2PO4)2}$$ is only weakly soluble into water : $$\ce{Ca(UO2)2(PO4)2 + 4 HCl -> Ca(H2PO4)2 (s) + 2 UO2Cl2(aq)}$$
But, as soon as some excess of $$\ce{HCl}$$ is added, both precipitates are redissolved according to the following global equation : $$\ce{Ca(UO2)2(PO4)2 + 6 HCl -> CaCl2(aq) + 2 UO2Cl2(aq) + 2H3PO4(aq)}$$ As the final substances are all soluble, they are dissociated into ions, so that the last equation should be better written, without $$\ce{Cl-}$$ ions : $$\ce{Ca(UO2)2(PO4)2 + 6 H+ -> Ca^{2+}(aq) + 2 UO2^{2+}(aq) + 2H3PO4(aq)}$$